# Complete lattice generated by a partitioning – finite meets

I conjectured certain formula for the complete lattice generated by a strong partitioning of an element of complete lattice. Now I have found a beautiful proof of a weaker statement than this conjecture. (Well, my proof works only in the case of distributive lattices, but the case of non-distributive lattices is outside of my research area.)

Let’s denote $R = \left\{ \bigcup{}^{\mathfrak{A}}X | X\in\mathscr{P}S \right\}$ where $S$ is a strong partitioning an element of the complete lattice $\mathfrak{A}$. Our conjecture is trivially equivalent to the statement that $R$ is closed under arbitrary meets and joins.

That $R$ is closed regarding any joins is obvious. To finish proving the conjecture we need to show that $R$ is closed under arbitrary meets. In this post I prove weaker result that $R$ is closed under finite meets.

I hope this finite case may serve as a model for the general infinite case. However it seems that generalizing it to infinite case is non-trivial.

Theorem Let $\mathfrak{A}$ is a distributive complete lattice and $S$ is a strong partitioning of some element of this lattice. Then $R$ is closed under finite meets.

Proof Let $X,Y\in\mathscr{P}S$.

Then $\bigcup^{\mathfrak{A}} X \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y = \bigcup^{\mathfrak{A}} ((X \cap Y) \cup (X \setminus Y)) \cap^{\mathfrak{A}}\bigcup^{\mathfrak{A}} Y = ( \bigcup^{\mathfrak{A}} (X \cap Y)\cup^{\mathfrak{A}} \bigcup^{\mathfrak{A}} (X \setminus Y))\cap^{\mathfrak{A}} \bigcup Y = ( \bigcup^{\mathfrak{A}} (X \cap Y)\cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y) \cup ( \bigcup^{\mathfrak{A}} (X\setminus Y) \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y) = (\bigcup^{\mathfrak{A}} (X \cap Y) \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}}Y) \cup^{\mathfrak{A}} 0 = \bigcup^{\mathfrak{A}} (X \cap Y)\cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y.$

Applying the formula $\bigcup^{\mathfrak{A}} X \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y = \bigcup^{\mathfrak{A}} (X \cap Y) \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y$ twice we get

$\bigcup^{\mathfrak{A}} X \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y = \bigcup^{\mathfrak{A}} (X \cap Y) \cap^{\mathfrak{A}} \bigcup (Y \cap (X \cap Y)) = \bigcup^{\mathfrak{A}} (X \cap Y) \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} (X \cap Y) = \bigcup^{\mathfrak{A}} (X \cap Y).$

But for any $A,B\in R$ exist $X,Y\in\mathscr{P}S$ such that $A=\bigcup^{\mathfrak{A}}X$ and $B=\bigcup^{\mathfrak{A}}Y$. So $A\cap^{\mathfrak{A}}B = \bigcup^{\mathfrak{A}} X \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y = \bigcup^{\mathfrak{A}} (X \cap Y) \in R$.